Multiple integral of min [duplicate]

How can this be proven? $$\int_0^1 \int_0^1...\int_0^1 min(x_1,x_2,...,x_n) dx_1dx_2...dx_n = \frac1{n+1}$$

I tried to split the last integral in $$\int_0^{min(x_1,x_2,...,x_{n-1})} x_ndx_n + \int_{min(x_1,x_2,...,x_{n-1})}^1 min(x_1,x_2,...,x_{n-1})dx_n$$

And then continue in this manner. So the function that remains to be integrated satisfies this recurrence relation $$f_n = (1-x)f_{n-1} + \int f_{n-1}$$

I then thought to use the following notation

$$I_n = \frac{ (-1)^{n+1}x^n }{n} + \frac{ (-1)^{n}x^{n-1} }{n-1}$$

If we compute the last integral, the n-1 integrals apply to $I_2$ If we apply another integral we get $I_3+I_2$ and so on, which will look like Pascal's triangle. But I can't go any further.

-

marked as duplicate by user1551, Claude Leibovici, Najib Idrissi, M Turgeon, mauMay 7 '14 at 12:22

Here is another opproach:

First consider the set $A=\{(t_1,\dots,t_n):t_1\le\dots\le t_n\}$. In this set, your integral becomes $$\int_0^1 \int_0^{t_n}...\int_0^{t_2} \int_0^{t_1} x_1 dx_1dx_2...dx_n = \frac1{(n+1)!}.$$

Since there are $n!$ sets like $A$ which are the union is $[0,1]^n$ then $$\int_0^1 \int_0^1...\int_0^1 min(x_1,x_2,...,x_n) dx_1dx_2...dx_n = n!\times \frac1{(n+1)!}=\frac1{n+1}.$$

-
Very nice! (+1) – Start wearing purple May 7 '14 at 8:48
Thank you! Very nice indeed. – user42768 May 7 '14 at 9:15
You're wellcome ;) – Jlamprong May 7 '14 at 9:31
This elegant solution deserves +1 upvote! – Sangchul Lee May 7 '14 at 9:42

First, write

\begin{align*} I &:= \int_{0}^{1} \cdots \int_{0}^{1} \min \{ x_{1}, \cdots, x_{n} \} \, dx_{n}\cdots dx_{1} \\ &= \int_{0}^{1} \cdots \int_{0}^{1} \int_{0}^{\min \{ x_{1}, \cdots, x_{n} \}} dt \, dx_{n}\cdots dx_{1}. \end{align*}

By noting that $t \leq \min \{x_{1}, \cdots, x_{n}\}$ is equivalent to $t \leq x_{1}, \cdots, t \leq x_{n}$, we find that $I$ is the volume of the region $\mathcal{W}$ defined by

$$\mathcal{W} = \{ (t, x_{1}, \cdots, x_{n}) : 0 \leq t \leq 1, t \leq x_{1} \leq 1, \cdots, t \leq x_{n} \leq 1 \}.$$

Thus by Fubini's Theorem we have

\begin{align*} I &= \int_{0}^{1} \int_{t}^{1} \cdots \int_{t}^{1} \, dx_{n}\cdots dx_{1} \, dt = \int_{0}^{1} (1-t)^{n} \, dt = \frac{1}{n+1}. \end{align*}

This solution can be encoded in the language of probability as follows: Let $U_{1}, \cdots, U_{n}$ be i.i.d. uniform random variables on $[0, 1]$. Then

\begin{align*} &\int_{0}^{1} \cdots \int_{0}^{1} \min \{ x_{1}, \cdots, x_{n} \} \, dx_{n}\cdots dx_{1} \\ &\qquad = \Bbb{E}( \min\{ U_{1}, \cdots, U_{n} \} ) \\ &\qquad = \int_{0}^{1} \Bbb{P}(\min\{ U_{1}, \cdots, U_{n} \} > x) \, dx \\ &\qquad = \int_{0}^{1} \Bbb{P}( U_{1} > x, \cdots, U_{n} > x) \, dx \\ &\qquad = \int_{0}^{1} \Bbb{P}( U_{1} > x)^{n} \, dx \qquad (\because \text{i.i.d.}) \\ &\qquad = \int_{0}^{1} (1 - x)^{n} \, dx = \frac{1}{n+1}. \end{align*}

-
I am having a hard time understanding where this equality comes from, since I haven't studied continuous random variables $$E(min(...)) = \int_0^1 P(min(...)>x) dx$$. Thank you! – user42768 May 7 '14 at 9:16
@user42768, Actually, that follows from Fubini's Theorem but I admit that it is not an easy observation. My calculus-version solution has the same content as in that solution, so you may ignore that part. – Sangchul Lee May 7 '14 at 9:41
So the proof to that equality is based on the "calculus" method of solving the problem? Thank you for your time. – user42768 May 7 '14 at 9:58
How did you apply Fubini's theorem? can someone enlight me on this? thx – mohlee Feb 6 '15 at 4:32

Use the fact that, $\min \{a,b\} = (a+b-|a-b|)/2$ for all $a,b \in \mathbb R$.

-
Could you please elaborate? – user42768 May 7 '14 at 8:24
Sure, apologies if the conversation stops suddenly. I will pick it up later. I think to see this, first just work on the case $n=2$. Then use the above expression to evaluate it. I will try this out myself now and post soon. – dcs24 May 7 '14 at 8:30
I already solved for n=2 using the method I described. Using an absolute value requires splitting of the integral, so it is similar to the way I thought to solve it. – user42768 May 7 '14 at 8:34
Using the absolute value you can use symmetry along the line x=y to remove it :-) that is how far I was getting. I have to stop at the minute, I will try to resume soon, but it seems you now have a solution below! – dcs24 May 7 '14 at 8:45